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As you know, in the initial position Chess960 the king lies between the two rooks. Here are the rules for castling.

For a starting position P, let n(P) be the minimal number of moves that White needs to legally castle kingside. How many positions are there such that n(P)=1?, n(P)=2, and so on?

There are cases where we can castle n the first move, namely when the King on f1 and the rook on g1. There are also other extreme cases with the king on b1 and the rook on c1, so then we have to make space.

Maybe some engines can solve this question.

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    If you can count you can work out the answer yourself. There are only 960 positions to consider. The clue is in the name.
    – Brian Towers
    Aug 29 '19 at 17:39
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    Yes, but 1. I am mathematician, so I can count :). 2. Since I am a mathematician, I am lazy so I would like to know whether there are general principles which would avoid to count. And how n(P) is distributed. Aug 29 '19 at 20:17
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    @DavideGiraudo : 1.I am mathematician, so I can count (...) 2. Since I am a mathematician, I am lazy : I could name several counterexamples for both claims !
    – Evargalo
    Aug 30 '19 at 10:17
  • Is n(P) the minimal number of moves before White can castle if he plays alone, or in any possible game (I.e. Black can help)?
    – Evargalo
    Aug 31 '19 at 6:32
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    Ah I missed this aspect. I am interested in the minimum in any possible game so like the example of your answer shows, black can help to make space by taking a pawn in the second row. In this way, they do not have to spend a time. Aug 31 '19 at 8:17
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Edit : this answer only deals with short castle (on the kingside), the figures have to be revised if you also consider long castle (on the queenside).

The three first steps:

There are 90 positions where n(P)=1

For White to be able to castle short at once, he needs to have Kf1, Rg1 in the initial position. Then there are 5 possible spots from the second rook, 3 for one bishop and 2 for the other one, and finally 3 remaining spots for the queen. The knights go to the last two spots.

There are 64 positions where n(P)=2

36 of them with Kf1, Ng1, Rh1: White can play 1.Nf3 or 1.Nh3 and 2.0-0

28 of them with Rg1, Nf1, Ke1: White can play 1.Ng3 or 1.Ne3 and 2.0-0

There are 236 positions where n(P)=3 please double-check this part

Here are the different configurations and the number of relevant positions:

xxxxxQKR 18
xxxxxBKR 36
xxxxxRKR 18
xxxxxKQR 18
xxxxxKBR 24
xxxxKNNR 8
xxxKNNRx 6
xxxKNRNx 6
xxxKRNNx 6
xxxxKBRx 24
xxxxKQRx 18
xxxxKRBx 36
xxxxKRQx 18

Total:236

Edit : removed 2 cases after checking the rules: the path to g1 must be open for the king (only the right-hand rook can be in the way)

Once we reach n(P)=4, we cannot ignore Black's play anymore: for instance, QNRNKBBR looks like it should have n=5, but actually it has n=4 because of the sequence 1.f4 b5 2.Bxa7 Qxg2 3.Bxg2 Nc6 4.0-0. Counting by hand becomes too fastidious and hazardous: a software program is needed.


On the other hand, after clarification by the OP, I believe 7 to be the maximum value of the function P->n(P).

If Black wasn't allowed to "help" we could build only 2 positions with n(P)=9; Here is one, the other is similar with RKBBNNQR on the first rank.

[FEN "rkqnnbbr/pppppppp/8/8/8/8/PPPPPPPP/RKQNNBBR w - - 0 1"]

In a series problem where White plays alone he would need 9 moves in order to castle. But since Black moves are taken into account, White actually can castle on move 7 from that diagram if the game goes something like 1.b3 d5 2.Qa3 Qe6 3.Nb2 Qxe2 4.Nd3 Qxf2 5.Bxf2 Nc6 6.Be2 0-0-0 7.0-0.

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  • Alternatively, Black can be a spoilsport and throw a Q in the works ;-) The question could be solved in both cases with problemist software and some patience (indeed, only 960 positions to check), but offhand I have no idea if e.g. Popeye is 960-compliant. Sep 22 at 8:57

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